为什么chan不会在goroution中陷入僵局？

Question

package main
func main() {
    ch1 := make(chan int)
    go pump(ch1) // pump hangs
}
func pump(ch chan int) {
    ch <- 1
    ch <- 2
}

Why is it that there is nothing wrong with writing a corruption? Blocking the main process will deadlock

Answer 1

Because the main() function exits and terminates the program while the pump() goroutine is blocked writing to the channel.

I think this code does what you expect:

package main
func main() {
    ch1 := make(chan int)
    done := make(chan struct{})
    go pump(ch1, done) // pump hangs
    <-done // wait for pump
}
func pump(ch chan int, done chan struct{}) {
    ch <- 1
    ch <- 2 // never reached
    close(done) // never reached
}

Output:

fatal error: all goroutines are asleep - deadlock!

goroutine 1 [chan receive]:
main.main()
    /tmp/sandbox778506014/main.go:7 +0xc0

goroutine 4 [chan send]:
main.pump(0x1043a080, 0x1043a0c0)
    /tmp/sandbox778506014/main.go:10 +0x40
created by main.main
    /tmp/sandbox778506014/main.go:6 +0xa0

Playground: https://play.golang.org/p/FTFPOBCm4G_0